Rotary Direct-Current Electric Motor With Mechanical Load
The electromechanical system consists of a rotary dc (direct-current) electric motor and load shaft. The dynamic model of this system consists of equations that describe the mechanical, electrical, and mechanical–electrical (coupled-field) behavior. Essentially, the electromechanical system sketched in Figure 1 is formed of a mobile part (the rotor armature), which rotates under the action of a magnetic field produced by the electrical circuit of an armature (the stator).
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| Figure 1: Schematic of a Rotary-Motion DC Motor With Load as an Electromagnetomechanical System. |
The electrical circuit is governed by Kirchhoff's second law, according to which
(10.78)
where the subscript a indicates the armature and vb is the back electromotive force (voltage). The mechanical part of the system is governed by the equation
(10.79)
where ma is the torque developed due to the stator-rotor interaction. It is also known that the following equations couple the mechanical and electrical fields:
(10.80)
with Km (measured in N·m/A) and Ke (measured in V·s/rad) being constants. The first Eq. (10.80) is similar to Eq. (10.72), which expressed the mechanical moment resulting from the interaction of an external magnetic field with the mobile arm carrying an electric current. The second Eq. (10.80) is similar to Eq. (10.73) that exemplified the back electromotive force being generated in a circuit with a current-carrying mobile arm in an external magnetic field.
When neglecting the armature inductance, which is La = 0 in Eq. (10.78), the following equation can be written by combining Eqs. (10.78) and (10.80)
(10.81)
and using the angular velocity ω(t) = dθ(t)/dt instead of the rotation angle θ(t). Eq. (10.81) indicates a linear relationship between the actuation torque ma and the angular velocity ω for a specified armature voltage—this relationship is plotted with solid line in Figure 2. The actuation torque–angular velocity curve generated by an armature voltage va1 intersects the coordinate axes at ω0—the free (no-load) angular velocity, when basically the dc motor shaft spins freely with no external load acting on it, and at ma,b—the block (stall) torque, where an external load torque equal to ma,b stalls the shaft rotation altogether. These two parameters are obtained from Eq. (10.81) as
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| Figure 2: Actuation and Load Torques Versus Shaft Angular Velocity in a Rotary DC Moto |
What's the easy way to measure a DC hobby motor's inductance?
When using simulations(SPICE), I need to model a DC motor or an electromagnet, relay ect. as an RL in series.
I can measure the R resistance by an ohmmeter. But I couldn't find an easy way to find out L value. Should I use a sinusoidal AC signal and measure the impedance from the phasor equations for that? But then I would need other equipment.
If you have a square wave generator and an oscilloscope then you can use the L/R time constant method. Hook the generator up to the coil in series with a resistor, and put the scope across the resistor. Adjust the resistor value and square wave frequency to get an exponential waveform like this:-
Now measure the time it takes for the voltage to drop to 37% of its peak value. This is the L/R time constant, T = L/R. R is the total resistance in the circuit (resistance of the coil + your resistor). You know T and R, so put those values into the formula and rearrange it to get L = T*R.
the DC resistance you can work out a scheme where the resistance will probably not make more than 10%-ish difference. Then you can estimate the inductance by just bluntly assuming the inductance is all you are measuring with a fixed voltage pulse. Of course the internal resistance will always make a difference and the line will never be perfectly straight and if you know all those modelling paradigms and understand the maths with all the e-power stuff, then you can use the resulting curved line to get a much more accurate prediction. But if you don't if you see a very strongly curved line, you need to adjust the the pulse time downward to limit the current built up, so that you stay in the mostly-linear region. (Or you can crank up the voltage, but be careful to check that the motor doesn't move).
If you use a fixed voltage the current through a perfect inductance is determined just by the voltage across it and the time that voltage has been across it, taken over the value of the inductor, like so:
I = (V*t)/L
To reiterate: This only counts for a fixed voltage, this is the result of an integration over constant voltage over a perfect inductance. Thus you need to make sure the internal DC resistance can be neglected (the line should be reasonably flat).
So you connect it like so:
By letting the scope plot (and record if possible) the signal of (Probe 2 - Probe 1) you see the voltage across the current measurement resistor. You then apply a voltage that will not make the motor spin (else you get back-EMF and all kinds of contact noise screwing up your results in more ways than one).
Select a MOSFET with a nice low R-on that is specifically made for switching on/off loads. Preferably inductive ones and even more preferably ones that are larger than yours (the less time and energy the MOSFET wastes, the better your measurement).
Then, if you turn on the MOSFET with a pulse signal after the motor has completely relaxed, you can see that current (voltage over the resistor) ramp up. When the pulse is 1s and you see a current of 1A at the end of it, after applying 1V, knowing the resistor is 0.1Ohm or less, you ignore the resistor and get:
L = (V*t)/I = (1*1)/1 = 1H Which would of course be ridiculous, but in example-land ridiculous is allowed.
Of course, if you are assuming the Rdc is negligible, you should make your current sense resistor just as small or smaller, else you will be adding more parasitic resistance.
The trick, usually, is to select a voltage high enough to let you get a good response with a reasonable pulse length, but that does only just not make anything actually move.
